A cylindrical gas container is closed at the top and open at the bottom.  If the iron plate of the top is 5/4 times as thick as the plate forming the cylindrical sides, the ratio of the radius to the height of the cylinder using minimum material for the same capacity is

Let  $r$ be the radius and  $h$ the height of the cylindrical gas container.  Also, let $k$  be the thickness of the plates forming the cylindrical sides. Therefore, the thickness of the plate forming the top will be  $5k/4.$
Capacity of the vessel = volume of cylinder
$=\pi r^{2}h=V$ (Given)    $\therefore h=V/\left(\pi x^2\right)\mathrm{.....}\left(1\right)$
Now, the volume $V_1$ of the iron plat used for construction of the container is given by
$$\begin{gathered} V_1=\pi {(r+k)}^2(h+5k/4)-\pi r^{2}h \\ \Rightarrow \frac{d{V}_1}{dr}=2\pi (r+k)(\frac{V}{\pi r^2}+\frac{5k}{4})+\pi {(r+k)}^2(\frac{V}{\pi }-2r^{-3})-0 \\ \therefore \frac{d{V}_1}{dr}=2Vk(r+k)\times (\frac{5\pi }{4V}-\frac{1}{r^3}) \end{gathered}$$

For maximum or minimum of  $V_1,d{V}_1/dr=0$ or  $r^3=[4V/\left(5\pi \right)].$
For this value of  $r,d^2{V}_1/d{r}^2$ is  $+ve.$
Hence,  $V_1$ is minimum when  $r={[4V/\left(5\pi \right)]}^{1/3}$
Now,  $x={[4V/\left(5\pi \right)]}^{1/3}$
or  $5\pi r^3=4V=4\pi r^{2}h$ or   $r/h=4/5$
Hence, the required ratio is 4 : 5.